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 Quote by Ben Niehoff This has nothing to do with it. Field lines work because the source-free Maxwell equations imply that the 2-form F is harmonic. Harmonic forms (in any dimension) have the distinction that they capture purely topological information. If you integrate a harmonic n-form over a closed n-surface, the result is either zero or non-zero, depending on whether the n-surface encloses some topological feature (for example, a 1-surface on a cylinder might wrap around the cylinder...or a 2-surface in R^3 might enclose a charge). Any n-surface that encloses the same set of topological features must give you the same result. You can think of this as a higher-dimensional analogue of contour integration. In fact, all analytic functions on the complex plane satisfy Laplace's equation, which is why contour integration works.
Just to clarify, by harmonic you mean: $$(d\delta+\delta d)F=0$$ where d is the exterior derivative and delta is the co-dfifferential?

Is this true? It seems since dF=0 by definition, then we need to show: $$d\delta F=-d(*d*F)=4\pi(d**J)=0$$

Is it true that $$dJ=0$$? I can't think of a reason for this...

EDIT: Wait, is that just conservation of charge? It certainly isn't the usual way of expressing it...and I can't seem think think clearly enough at this late hour to figure this out...lol...